Question

Prove that for an ideal-gas reaction, `(dlnK_C^@)/(dT) = (DeltaU^@)/(RT^2)`?

I found this question particularly difficult, but I figured it out and wanted to share.

Answer 1

For an ideal gas reaction, we begin with the definition of `K_C^@`

`K_C^@ = K_P^@((P^@)/(RTc^@))^(Deltan)`

and differentiate `lnK_C^@` with respect to `T`. We would obtain, noting the usage of the product rule on `K_P^@ = e^(-DeltaG^@/(RT))` (since `DeltaG^@ = DeltaG^@(T)`):

`(dlnK_C^@)/(dT)`

`= d/(dT)[ln{e^(-(DeltaG^@)/(RT))((P^@)/(RTc^@))^(Deltan)}]`

`= d/(dT)[-(DeltaG^@)/(RT) + Deltanln((P^@)/(RTc^@))]`

`= d/(dT)[-(DeltaG^@)/(RT)] + Deltan(cancel(RTc^@)/cancel(P^@))*-cancel(P^@)/(cancel(Rc^@)T^cancel(2))`

`= stackrel("Product Rule")overbrace((DeltaG^@)/(RT^2) - 1/(RT)(dDeltaG^@)/(dT)) - (Deltan)/T`

By definition, since `DeltaG^@` is defined for a fixed `"1 bar"` pressure, we can refer to the Maxwell relation

`dG^@ = -S^@dT + VdP^@`

or

`dDeltaG^@ = -DeltaS^@dT + DeltaVdP^@`

and acquire

`((delDeltaG^@)/(delT))_(P^@) = (dDeltaG^@)/(dT) = -DeltaS^@`

Then we proceed to acquire the result by noting that `DeltaG^@ = DeltaH^@ - TDeltaS^@`, and that for an ideal gas, `DeltaH^@ = DeltaU^@ + Delta(PV) = DeltaU^@ + DeltanRT`:

`=> (DeltaG^@)/(RT^2) + (DeltaS^@)/(RT) - (Deltan)/T = (DeltaG^@)/(RT^2) + (TDeltaS^@)/(RT^2) - (DeltanRT)/(RT^2)`

`= (DeltaG^@ + TDeltaS^@ - DeltanRT)/(RT^2) = (DeltaH^@ - cancel(TDeltaS^@ + TDeltaS^@) - DeltanRT)/(RT^2)`

`= (DeltaU^@ + Delta(PV) - DeltanRT)/(RT^2) = (DeltaU^@ + cancel(DeltanRT) - cancel(DeltanRT))/(RT^2)`

`= color(blue)((DeltaU^@)/(RT^2))`